Section 2.1 Solving Linear Equations 61 Objective 7 Examples A problem and an incorrect solution are given. Provide the correct solution and an explanation of the error. 7a. Solve -3m + 5=-7. Incorrect Solution Correct Solution and Explanation -3m + 5=-7 -3m + - 5=-7 - 5 -3m=-12 -3m + 3=-12 + 3 m=-9 The solution set is 5-96. 7b. Solve x 4 + 5 2 = 3. Incorrect Solution Correct Solution and Explanation x 5 The error was made in not multiplying + = 3 4 2 each side by the LCD. x 4 + 5 2 = 3 8¢ x 4 + 5 2 ≤ = 8(3) 8¢ x 4 ≤ + 8¢ 5 2 ≤ = 24 2x + 20 = 24 2x + 20 - 20 = 24 - 20 2x = 4 x = 2 The solution set is 526. 8¢ x 4 + 5 2 ≤ = 3 8¢ x 4 ≤ + 5 2 = 3 2x + 20 = 3 2x + 20 - 20 = - 20 2x=-x=- 17 2 The solution set is e- 2 . 7c. Solve 4 - 3(2x - 4) = 4(x + 2) + 8. Incorrect Solution Correct Solution and Explanation 4 - 3(2x - 4) = 4(x + 2) + 8 In the last step, we should divide each side by 10 to make the coefficient of x 1. 4 - 3(2x - 4) = 4(x + 2) + 8 4 - 6x + 12 = 4x + 8 + 8 -6x + 16 = 4x + 16 16 = 10x + 16 0 = 10x 0 10x = 10 10 0 = x The solution set is 506. 4 - 6x 12 = 4x + 8 + 8 -6x + = 4x + 16 16 10x + 16 0 = 10x The solution set is . To make the coefficient of m 1, we must divide by -3, not add -3. Keep in mind that -3m + 3 can’t be simplified but -3m = m. -3 -3m + 5=-7 -3m + 5 - 5=-7 - 5 -3m=-12 -3m -3 = -12 -3 m = 4 The solution set is 5-46. 3 5 3m 3 = - tion s 8¢ ≤ 0 3 17 i e 17 f 8 + 1 16 = 1 0x olu
hendricks_intermediate_algebra_1e_ch1_3
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